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A Collection of Limits

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Contents
1 Short theoretical introduction

1

2 Problems

12

3 Solutions

23

2

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Chapter 1

Short theoretical
introduction
Consider a sequence of real numbers (an )n≥1 , and l ∈ R. We’ll say that l
represents the limit of (an )n≥1 if any neighborhood of l contains all the terms of
the sequence, starting from a certain index. We write this fact as lim an = l,
n→∞
or an → l.
We can rewrite the above definition into the following equivalence:
lim an = l ⇔ (∀)V ∈ V(l), (∃)nV ∈ N∗ such that (∀)n ≥ nV ⇒ an ∈ V .

n→∞

One can easily observe from this definition that if a sequence is constant then
it’s limit is equal with the constant term.
We’ll say that a sequence of real numbers (an )n≥1 is convergent if it has limit
and lim an ∈ R, or divergent if it doesn’t have a limit or if it has the limit
n→∞
equal to ±∞.
Theorem: If a sequence has limit, then this limit is unique.
Proof: Consider a sequence (an )n≥1 ⊆ R which has two different limits l , l ∈ R.
It follows that there exist two neighborhoods V ∈ V(l ) and V ∈ V(l ) such
that V ∩ V = ∅. As an → l ⇒ (∃)n ∈ N∗ such that (∀)n ≥ n ⇒ an ∈ V .
Also, since an → l ⇒ (∃)n ∈ N∗ such that (∀)n ≥ n ⇒ an ∈ V . Hence
(∀)n ≥ max{n , n } we have an ∈ V ∩ V = ∅.
Theorem: Consider a sequence of real numbers (an )n≥1 . Then we have:
(i) lim an = l ∈ R ⇔ (∀)ε > 0, (∃)nε ∈ N∗ such that (∀)n ≥ nε ⇒ |an − l| < ε.
n→∞

1

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2

A Collection of Limits

(ii) lim an = ∞ ⇔ (∀)ε > 0, (∃)nε ∈ N∗ such that (∀)n ≥ nε ⇒ an > ε.
n→∞

(iii) lim an = −∞ ⇔ (∀)ε > 0, (∃)nε ∈ N∗ such that (∀)n ≥ nε ⇒ an < −ε
n→∞

Theorem: Let (an )n≥1 a sequence of real numbers.
1. If lim an = l, then any subsequence of (an )n≥1 has the limit equal to l.
n→∞

2. If there exist two subsequences of (an )n≥1 with different limits, then the
sequence (an )n≥1 is divergent.
3. If there exist two subsequences of (an )n≥1 which cover it and have a common
limit, then lim an = l.
n→∞

Definition: A sequence (xn )n≥1 is a Cauchy sequence if (∀)ε > 0, (∃)nε ∈ N
such that |xn+p − xn | < ε, (∀)n ≥ nε , (∀)p ∈ N.
Theorem: A sequence of real numbers is convergent if and only if it is a Cauchy
sequence.
Theorem: Any increasing and unbounded sequence has the limit ∞.
Theorem: Any increasing and bounded sequence converge to the upper bound
of the sequence.
Theorem: Any convergent sequence is bounded.
Theorem(Cesaro lemma): Any bounded sequence of real numbers contains
at least one convergent subsequence.
Theorem(Weierstrass theorem): Any monotonic and bounded sequence is
convergent.
Theorem: Any monotonic sequence of real numbers has limit.
Theorem: Consider two convergent sequences (an )n≥1 and (bn )n≥1 such that
an ≤ bn , (∀)n ∈ N∗ . Then we have lim an ≤ lim bn .
n→∞

n→∞

Theorem: Consider a convergent sequence (an )n≥1 and a real number a such
that an ≤ a, (∀)n ∈ N∗ . Then lim an ≤ a.
n→∞

Theorem: Consider a convergent sequence (an )n≥1 such that lim an = a.
n→∞

Them lim |an | = |a|.
n→∞

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Short teoretical introduction

3

Theorem: Consider two sequences of real numbers (an )n≥1 and (bn )n≥1 such
that an ≤ bn , (∀)n ∈ N∗ . Then:
1. If lim an = ∞ it follows that lim bn = ∞.
n→∞

n→∞

2. If lim bn = −∞ it follows that lim an = −∞.
n→∞

n→∞

Limit operations:
Consider two sequences an and bn which have limit. Then we have:
1. lim (an + bn ) = lim an + lim bn (except the case (∞, −∞)).
n→∞

n→∞

n→∞

2. lim (an · bn ) = lim an · lim bn (except the cases (0, ±∞)).
n→∞

n→∞

n→∞

lim an
an
= n→∞ (except the cases (0, 0), (±∞, ±∞)).
n→∞ bn
lim bn

3. lim

n→∞

lim bn

4. lim abnn = ( lim an )n→∞
n→∞

n→∞

(except the cases (1, ±∞), (∞, 0), (0, 0)).

5. lim (logan bn ) = log lim a ( lim bn ).
n n→∞
n→∞
n→∞

Trivial consequences:
1. lim (an − bn ) = lim an − lim bn ;
n→∞

n→∞

n→∞

2. lim (λan ) = λ lim an (λ ∈ R);
n→∞

3. lim

n→∞

n→∞

k

an =

k

lim an (k ∈ N);

n→∞

Theorem (Squeeze theorem): Let (an )n≥1 , (bn )n≥1 , (cn )n≥1 be three sequences of real numbers such that an ≤ bn ≤ cn , (∀)n ∈ N∗ and lim an =
n→∞

lim cn = l ∈ R. Then lim bn = l.

n→∞

n→∞

Theorem: Let (xn )n≥1 a sequence of real numbers such that lim (xn+1 −xn ) =
n→∞

α ∈ R.
1. If α > 0, then lim xn = ∞.
n→∞

2. If α < 0, then lim xn = −∞.
n→∞

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4

A Collection of Limits

Theorem (Ratio test): Consider a sequence of real positive numbers (an )n≥1 ,
an+1
for which l = lim
∈ R.
n→∞ an
1. If l < 1 then lim an = 0.
n→∞

2. If l > 1 then lim an = ∞.
n→∞

an+1
,
Proof: 1. Let V = (α, β) ∈ V(l) with l < β < 1. Because l = lim
n→∞ an
a
n+1
there is some n0 ∈ N∗ such that (∀)n ≥ n0 ⇒
∈ V , hence (∀)n ≥ n0 ⇒
an
an+1
< 1. That means starting from the index n0 the sequence (an )n≥1 is
an
strictly decreasing. Since the sequence is strictly decreasing and it contains
only positive terms, the sequence is bounded. Using Weierstrass Theorem, it
follows that the sequence is convergent. We have:
an+1 =

an+1
an+1
· an ⇒ lim an+1 = lim
· lim an
n→∞
n→∞
an
an n→∞

which is equivalent with:
lim an (1 − l) = 0

n→∞

which implies that lim an = 0.
n→∞

1
bn+1
1
we have lim
=
< 1, hence lim bn = 0 which
n→∞ bn
n→∞
an
l
implies that lim an = ∞.
2. Denoting bn =
n→∞

Theorem: Consider a convergent sequence of real non-zero numbers (xn )n≥1
xn
such that lim n
− 1 ∈ R∗ . Then lim xn = 0.
n→∞
n→∞
xn−1
Theorem(Cesaro-Stolz lemma): 1. Consider two sequences (an )n≥1 and
(bn )n≥1 such that:
(i) the sequence (bn )n≥1 is strictly increasing and unbounded;
an+1 − an
= l exists.
n→∞ bn+1 − bn

(ii) the limit lim

Then the sequence

an
bn

an
= l.
n→∞ bn

is convergent and lim
n≥1

Proof: Let’s consider the case l ∈ R and assume (bn )n≥1 is a strictly increasing
sequence, hence lim bn = ∞. Now let V ∈ V(l), then there exists α > 0 such
n→∞

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Short teoretical introduction

5

that (l − α, l + α) ⊆ V . Let β ∈ R such that 0 < β < α. As lim

n→∞

exists k ∈ N∗ such that (∀)n ≥ k ⇒

an
= l, there
bn

an+1 − an
∈ (l − β, l + β), which implies
bn+1 − bn

that:
(l − β)(bn+1 − bn ) < an+1 − an < (l + β)(bn+1 − bn ), (∀)n ≥ k
Now writing this inequality from k to n − 1 we have:
(l − β)(bk+1 − bk ) < ak+1 − ak < (l + β)(bk+1 − bk )
(l − β)(bk+2 − bk+1 ) < ak+2 − ak+1 < (l + β)(bk+2 − bk+1 )
...
(l − β)(bn − bn−1 ) < an − an−1 < (l + β)(bn − bn−1 )
Summing all these inequalities we find that:
(l − β)(bn − bk ) < an − ak < (l + β)(bn − bk )
As lim bn = ∞, starting from an index we have bn > 0. The last inequality
n→∞
rewrites as:
(l − β) 1 −
⇔ (l − β) +

bk
bn

<

an
ak
bk

< (l + β) 1 −
bn
bn
bn

ak + (β − l)bk
an
ak − (β + l)bk
<
bn
bn
bn

As
ak + (β − l)bk
ak − (β + l)bk
= lim
=0
n→∞
n→∞
bn
bn
lim

there exists an index p ∈ N∗ such that (∀)n ≥ p we have:
ak + (β − l)bk ak − (β + l)bk
,
∈ (β − α, α − β)
bn
bn
We shall look for the inequalities:
ak + (β − l)bk
>β−α
bn
and
ak − (β + l)bk
<α−β
bn

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6

A Collection of Limits

Choosing m = max{k, p}, then (∀)n ≥ m we have:
l−α<

an
bn

an
an
∈ V ⇒ lim
= l. It remains to prove the theorem
n→∞ bn
bn
when l = ±∞, but these cases can be proven analogous choosing V = (α, ∞)
and V = (−∞, α), respectively.
which means that

2. Let (xn )n≥1 and (yn )n≥1 such that:
(i) lim xn = lim yn = 0, yn = 0, (∀)n ∈ N∗ ;
n→∞

n→∞

(ii) the sequence (yn )n≥1 is strictly decreasing;
xn+1 − xn
= l ∈ R.
n→∞ yn+1 − yn

(iii) the limit lim

xn
yn

Then the sequence

has a limit and lim

n→∞

n≥1

xn
= l.
yn

xn
xn+1 − xn
= lim
,
n→∞ yn+1 − yn
yn
and if the limit we arrive to belongs to R, then the application of Cesaro-Stolz
lemma is valid.
Remark: In problem’s solutions we’ll write directly lim

n→∞

Trivial consequences:
1. Consider a sequence (an )n≥1 of strictly positive real numbers for which exists
an+1
lim
= l. Then we have:
n→∞ an

an+1
lim n an = lim
n→∞ an
n→∞
Proof: Using Cesaro-Stolz theorem we have:

lim (ln

n→∞

n

ln an
ln an+1 − ln an
= lim
= lim ln
n→∞ n
n→∞ (n + 1) − n
n→∞

an ) = lim

an+1
an

Then:
lim

n→∞

n

an = lim eln
n→∞

n

an

lim (ln

= en→∞

n

an )

= eln l = l

2. Let (xn )n≥1 a sequence of real numbers which has limit. Then:
lim

n→∞

x1 + x2 + . . . + xn
= lim xn
n→∞
n

= ln l

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Short teoretical introduction

7

3. Let (xn )n≥1 a sequence of real positive numbers which has limit. Then:
lim

n

n→∞

x1 x2 . . . xn = lim xn
n→∞

Theorem (Reciprocal Cesaro-Stolz): Let (xn )n≥1 and (yn )n≥1 two sequences of real numbers such that:
(i) (yn )n≥1 is strictly increasing and unbounded;
(ii) the limit lim

xn
= l ∈ R;
yn

(iii) the limit lim

yn
∈ R+ \{1}.
yn+1

n→∞

n→∞

Then the limit lim

n→∞

xn+1 − xn
exists and it is equal to l.
yn+1 − yn

Theorem (exponential sequence): Let a ∈ R. Consider the sequence xn =
an , n ∈ N∗ .
1. If a ≤ −1, the sequence is divergent.
2. If a ∈ (−1, 1), then lim xn = 0.
n→∞

3. If a = 1, then lim xn = 1.
n→∞

4. If a > 1, then lim xn = ∞.
n→∞

Theorem (power sequence): Let a ∈ R. Consider the sequence xn = na , n ∈
N∗ .
1. If a < 0, then lim xn = 0.
n→∞

2. If a = 0, then lim xn = 1.
n→∞

3. If a > 0, then lim xn = ∞.
n→∞

Theorem (polynomial sequence): Let an = ak nk + ak−1 nk−1 + . . . + a1 n +
a0 , (ak = 0).
1. If ak > 0, then lim an = ∞.
n→∞

2. If ak < 0, then lim an = −∞.
n→∞

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8

A Collection of Limits

Theorem: Let bn =

ak nk + ak−1 nk−1 + . . . + a1 n + a0
, (ak = 0 = bp ).
bp np + bp−1 np−1 + . . . + b1 n + b0

1. If k < p, then lim bn = 0.
n→∞

2. If k = p, then lim bn =

ak
.
bp

3. If k > p, then lim bn =

ak
· ∞.
bp

n→∞

n→∞

Theorem: The sequence an =

1+

n

1
n

, n ∈ N∗ is a strictly increasing and

bounded sequence and lim an = e.
n→∞

Theorem: Consider a sequence (an )n≥1 of real non-zero numbers such that
1
lim an = 0. Then lim (1 + an ) an = e.
n→∞

n→∞

Proof: If (bn )n≥1 is a sequence of non-zero positive integers such that lim bn =
1
1+
bn

n→∞
n

bn

1
∞, we have lim
= e, it
= e. Let ε > 0. From lim 1 +
n→∞
n→∞
n
n
1
follows that there exists nε ∈ N∗ such that (∀)n ≥ nε ⇒ 1 +
− e < ε.
n

Also, since lim bn = ∞, there exists nε ∈ N such that (∀)n ≥ nε ⇒ bn >
n→∞

nε . Therefore there exists nε = max{nε , nε } ∈ N∗ such that (∀)n ≥ nε ⇒
b
b
1 n
1 n
1+
− e < ε. This means that: lim 1 +
= e. The same
n→∞
bn
bn
property is fulfilled if lim bn = −∞.
n→∞

1
n→∞
n→∞
cn
e. We can assume that cn > 1, (∀)n ∈ N∗ . Let’s denote dn = cn ∈ N∗ . In
this way (dn )n≥1 is sequence of positive integers with lim dn = ∞. We have:
If (cn )n≥1 is a sequence of real numbers such that lim cn = ∞, then lim

1+

n→∞

dn ≤ cn < dn + 1 ⇒

1
1
1
<

dn + 1
cn
dn

Hence it follows that:

1+

1
dn + 1

Observe that:

d

<
n

1+

1
cn

dn

1+

1
cn

cn

<

1+

1
cn

dn +1

1+

1
dn

dn +1

cn

=

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Short teoretical introduction

lim

n→∞

1+

1
dn + 1

9

dn

= lim

n→∞

1+

dn +1

1
dn + 1

1
dn + 1

· 1+

−1

=e

and
lim

n→∞

1+

dn +1

1
dn

= lim

n→∞

dn

1
dn

1+

· 1+

Using the Squeeze Theorem it follows that lim

1+

n→∞

1
dn

1
cn

=e

cn

= e. The same

property is fulfilled when lim cn = −∞.
n→∞

Now if the sequence (an )n≥1 contains a finite number of positive or negative
terms we can remove them and assume that the sequence contains only positive
1
we have lim xn = ∞. Then we have
terms. Denoting xn =
n→∞
an
xn
1
1
lim (1 + an ) an = lim 1 +
=e
n→∞
n→∞
xn
If the sequence contains an infinite number of positive or negative terms, the
1
same fact happens for the sequence (xn )n≥1 with xn =
, (∀)n ∈ N∗ . Let’s
an
denote by (an )n≥1 the subsequence of positive terms , and by (an )n≥1 the subse1
1
quence of negative terms. Also let cn =
, (∀)n ∈ N∗ and cn =
, (∀)n ∈ N∗ .
an
an
Then it follows that lim cn = ∞ and lim cn = −∞. Hence:
n→∞

n→∞

1
cn

cn

1+

1
cn

cn

1+

1

lim (1 + an ) an = lim

n→∞

n→∞

=e

and
1

lim (1 + an ) an = lim

n→∞

n→∞

=e

1

Then it follows that: lim (1 + an ) an = e.
n→∞

Consequence: Let (an )n≥1 , (bn )n≥1 two sequences of real numbers such that
an = 1, (∀)n ∈ N∗ , lim an = 1 and lim bn = ∞ or lim bn = −∞. If there
n→∞

n→∞

exists lim (an − 1)bn ∈ R, then we have lim
n→∞

n→∞

abnn

=e

n→∞
lim (an −1)bn

n→∞

n

Theorem: Consider the sequence (an )n≥0 defined by an =
k=0

lim an = e.

n→∞

.
1
. We have
k!

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10

A Collection of Limits

Theorem: Let (cn )n≥1 , a sequence defined by
1 1
1
+ + . . . + − ln n, n ≥ 1
2 3
n
Then (cn )n≥1 is strictly decreasing and bounded, and lim cn = γ, where γ is
n→∞
the Euler constant.
cn = 1 +

Recurrent sequences
A sequence (xn )n≥1 is a k-order recurrent sequence, if it is defined by a formula
of the form
xn+k = f (xn , xn+1 , . . . , nn+k−1 ), n ≥ 1
with given x1 , x2 , . . . , xk . The recurrence is linear if f is a linear function.
Second order recurrence formulas which are homogoeneus, with constant coefficients, have the form xn+2 = αxn+1 + βxn , (∀)n ≥ 1 with given x1 , x2 , α, β.
To this recurrence formula we attach the equation r2 = αr + β, with r1 , r2 as
solutions.
If r1 , r2 ∈ R and r1 = r2 , then xn = Ar1n +Br2n , where A, B are two real numbers,
usually found from the terms x1 , x2 . If r1 = r2 = r ∈ R, then xn = rn (A + nB)
and if r1 , r2 ∈ R, we have r1 , r2 = ρ(cos θ + i sin θ) so xn = ρn (cos nθ + i sin nθ).
Limit functions
Definition: Let f : D → R (D ⊆ R) and x0 ∈ R and accumulation point
of D. We’ll say that l ∈ R is the limit of the function f in x0 , and we write
lim f (x) = l, if for any neightborhood V of l, there is a neighborhood U of x0 ,
x→x0

such that for any x ∈ D ∩ U \{x0 }, we have f (x) ∈ V.
Theorem: Let f : D → R (D ⊂ R) and x0 an accumulation point of D. Then
lim f (x) = l (l, x0 ∈ R) if and only if (∀)ε > 0, (∃)δε > 0, (∀)x ∈ D\{x0 }
x→x0

such that |x − x0 | < δε ⇒ |f (x) − l| < ε.
If l = ±∞, we have:
lim f (x) = ±∞ ⇔ (∀)ε > 0, (∃)δε > 0, (∀)x ∈ D\{x0 } such that |x−x0 | < δε ,

x→x0

we have f (x) > ε (f (x) < ε).
Theorem: Let f : D ⊂ R ⇒ R and x0 an accumulation point of D. Then
lim f (x) = l (l ∈ R, x0 ∈ R), if and only if (∀)(xn )n≥1 , xn ∈ D\{x0 }, xn →
x→x0

x0 , we have lim f (xn ) = l.
n→∞

One-side limits

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Short teoretical introduction

11

Definition: Let f : D ⊆ R → R and x0 ∈ R an accumulation point of D. We’ll
say that ls ∈ R (or ld ∈ R) is the left-side limit (or right-side limit) of f in x0 if
for any neigborhood V of ls (or ld ), there is a neighborhood U of x0 , such that
for any x < x0 , x ∈ U ∩ D\{x0 } (x > x0 respectively), f (x) ∈ V.
We write ls = lim f (x) = f (x0 − 0) and ld = lim f (x) = f (x0 + 0).
x → x0
x>x0

x → x0
x
Theorem: Let f : D ⊆ R → R and x0 ∈ R an accumulation point of the sets
(−∞, x0 ) ∩ D and (x0 , ∞) ∩ D. Then f has the limit l ∈ R if and only if f has
equal one-side limits in x0 .
Remarkable limits
If lim f (x) = 0, then:
x→x0

1. lim

sin f (x)
= 1;
f (x)

2. lim

tan f (x)
= 1;
f (x)

3. lim

arcsin f (x)
= 1;
f (x)

4. lim

arctan f (x)
= 1;
f (x)

x→x0

x→x0

x→x0

x→x0

1
f
(x)
5. lim (1 + f (x))
=e
x→x0

6. lim

ln(1 + f (x))
= 1;
f (x)

7. lim

af (x) − 1
= ln a (a > 0);
f (x)

8. lim

(1 + f (x))r − 1
= r (r ∈ R);
f (x)

x→x0

x→x0

x→x0

If lim f (x) = ∞, then:
x→x0

1
f (x)

f (x)

9. lim

1+

10. lim

ln f (x)
= 0;
f (x)

x→x0

x→x0

= e;

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Chapter 2

Problems
1. Evaluate:
3

lim

n→∞

n3 + 2n2 + 1 −

3

n3 − 1

2. Evaluate:

3
5x + 2 + 2
lim √
x→−2
3x + 10 − 2
n

3n2 + 9n
, (∀)n ≥ 1.
2
k=1
Prove that this sequence is an arithmetical progression and evaluate:

3. Consider the sequence (an )n≥1 , such that

1
n→∞ nan

ak =

n

lim

ak
k=1

1
(an +
3
2
an−1 + b), where 0 ≤ b ≤ 1. Prove that the sequence is convergent and evaluate
lim an .
4. Consider the sequence (an )n≥1 such that a1 = a2 = 0 and an+1 =

n→∞

5. Consider a sequence of real numbers (xn )n≥1 such that x1 = 1 and xn =
1
2xn−1 + , (∀)n ≥ 2. Evaluate lim xn .
n→∞
n
6. Evaluate:
lim

n→∞

n

4
5

n

+ n2 sinn

7. Evaluate:
12

π
π
+ cos 2nπ +
6
n

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Problems

13

n

lim

n→∞

k=1

k! · k
(n + 1)!

8. Evaluate:
lim

n→∞

1−

1
22

1−

1
32

· ... · 1 −

1
n2

9. Evaluate:
lim

n

n→∞

33n (n!)3
(3n)!

10. Consider a sequence of real positive numbers (xn )n≥1 such that (n+1)xn+1 −
nxn < 0, (∀)n ≥ 1. Prove that this sequence is convergent and evaluate it’s
limit.
11. Find the real numbers a and b such that:
3

lim

1 − n3 − an − b = 0

n→∞

12. Let p ∈ N and α1 , α2 , ..., αp positive distinct real numbers. Evaluate:
lim

n

n→∞

α1n + α2n + . . . + αpn

13. If a ∈ R∗ , evaluate:
lim

x→−a

cos x − cos a
x2 − a2

14. If n ∈ N∗ , evaluate:
ln(1 + x + x2 + . . . + xn )
x→0
nx
lim

15. Evaluate:
n

lim

n→∞

n2 + n −
k=1

2k 3 + 8k 2 + 6k − 1
k 2 + 4k + 3

16. Find a ∈ R such that:
lim

x→0

1 − cos ax
sin x
= lim
x→π π − x
x2

17. Evaluate:

3
lim

x→1

18. Evaluate:

x2 + 7 − x + 3
x2 − 3x + 2

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14

A Collection of Limits

2n2 + n − λ

lim

n→∞

2n2 − n

where λ is a real number.
19. If a, b, c ∈ R, evaluate:

lim a x + 1 + b x + 2 + c x + 3

x→∞

20. Find the set A ⊂ R such that ax2 + x + 3 ≥ 0, (∀)a ∈ A, (∀)x ∈ R. Then
for any a ∈ A, evaluate:
ax2 + x + 3

x+1−

lim

x→∞

21. If k ∈ R, evaluate:
n

n+1

lim nk

n→∞

n+2
n+3

22. If k ∈ N and a ∈ R+ \{1}, evaluate:
n−1

n

1

lim nk (a n − 1)

n→∞

n+1
n+2

23. Evaluate:
n

lim

n→∞

k=1

1
n2 + k

24. If a > 0, p ≥ 2, evaluate:
n

p

lim

n→∞

k=1

1
np + ka

25. Evaluate:
lim

n!

n→∞

(1 +

12 )(1

+

22 )

· . . . · (1 + n2 )

26. Evaluate:
2

lim

n→∞

2n − 3
2n2 − n + 1

n2 − 1
n

27. Evaluate:
lim

x→0

1 + sin2 x − cos x

1 − 1 + tan2 x

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Problems

15

28. Evaluate:

x+ x

x− x

lim

x→∞

x

29. Evaluate:
1

lim (cos x) sin x

x→0
x>0

30. Evaluate:
1

lim (ex + sin x) x

x→0

31. If a, b ∈ R∗+ , evaluate:
lim

n→∞

a−1+
a

n

n

b

32. Consider a sequence of real numbers (an )n≥1 defined by:

1
if n ≤ k, k ∈ N∗

k
k
(n
+
1)

n
an =
if n > k

n
k−1

i)Evaluate lim an .
n→∞

n

k · lim an , evaluate:

ii)If bn = 1 +
k=1

n→∞

lim

n→∞

b2n
bn−1 bn+1

n

33. Consider a sequence of real numbers (xn )n≥1 such that xn+2 =
N∗ . If x1 ≤ x2 ,

xn+1 + xn
, (∀)n ∈
2

i)Prove that the sequence (x2n+1 )n≥0 is increasing, while the sequence (x2n )n≥0
is decreasing;
ii)Prove that:
|xn+2 − xn+1 | =

|x2 − x1 |
, (∀)n ∈ N∗
2n

iii)Prove that:
2xn+2 + xn+1 = 2x2 + x1 , (∀)n ∈ N∗
iv)Prove that (xn )n≥1 is convergent and that it’s limit is

x1 + 2x2
.
3

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16

A Collection of Limits

34. Let an , bn ∈ Q such that (1 +
an
lim
.
n→∞ bn

2)n = an + bn 2, (∀)n ∈ N∗ . Evaluate

35. If a > 0, evaluate:
(a + x)x − 1
x→0
x
lim

36. Consider a sequence of real numbers (an )n≥1 such that a1 =

3
and an+1 =
2

a2n − an + 1
. Prove that (an )n≥1 is convergent and find it’s limit.
an

37. Consider a sequence of real numbers (xn )n≥1 such that x0 ∈ (0, 1) and
xn+1 = xn − x2n + x3n − x4n , (∀)n ≥ 0. Prove that this sequence is convergent
and evaluate lim xn .
n→∞

38. Let a > 0 and b ∈ (a, 2a) and a sequence x0 = b, xn+1 = a+ xn (2a − xn ), (∀)n ≥
0. Study the convergence of the sequence (xn )n≥0 .
39. Evaluate:
n+1

lim

n→∞

arctan
k=1

1
2k 2

40. Evaluate:
n

lim

n→∞

k=1

k
+1

4k 4

41. Evaluate:
n

lim

n→∞

k=1

1 + 3 + 32 + . . . + 3k
5k+2

42. Evaluate:
n

lim

n→∞

i

n+1−
i=2 k=2

k−1
k!

43. Evaluate:
11 + 22 + 33 + . . . + nn
n→∞
nn
lim

44. Consider the sequence (an )n≥1 such that a0 = 2 and an−1 − an =
Evaluate lim ((n + 1)! ln an ).
n→∞

n
.
(n + 1)!

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Problems

17

45. Consider a sequence of real numbers (xn )n≥1 with x1 = a > 0 and xn+1 =
x1 + 2x2 + 3x3 + . . . + nxn
, n ∈ N∗ . Evaluate it’s limit.
n
n

46. Using lim

n→∞

k=1

π2
1
=
, evaluate:
2
k
6
n

1
(2k − 1)2

lim

n→∞

k=1

47. Consider the sequence (xn )n≥1 defined by x1 = a, x2 = b, a < b and
xn−1 + λxn−2
xn =
, n ≥ 3, λ > 0. Prove that this sequence is convergent and
1+λ
find it’s limit.
48. Evaluate:
n
lim √
n
n!

n→∞

49. Consider the sequence (xn )n≥1 defined by x1 = 1 and xn =

1
, n≥
1 + xn−1

2. Prove that this sequence is convergent and evaluate lim xn .
n→∞

50. If a, b ∈ R∗ , evaluate:
lim

x→0

51. Let f : R → R, f (x) =

ln(cos ax)
ln(cos bx)

{x}
x

if x ∈ Q
. Find all α ∈ R for which
if x ∈ R\Q

x
x

if x ∈ Q
. Find all α ∈ R for which
if x ∈ R\Q

lim f (x) exists.

x→α

52. Let f : R → R, f (x) =
lim f (x) exists.

x→α

53. Let (xn )n≥1 be a sequence of positive real numbers such that x1 > 0 and
a
3xn = 2xn−1 + 2 , where a is a real positive number. Prove that xn is
xn−1
convergent and evaluate lim xn .
n→∞

54. Consider a sequence of real numbers (an )n≥1 such that a1 = 12 and an+1 =
3
an 1 +
. Evaluate:
n+1
n

lim

n→∞

k=1

1
ak

diendantoanhoc.net [VMF]

18

A Collection of Limits

55. Evaluate:

lim

n→∞

n

n
n2 + 1

56. If a ∈ R, evaluate:
n

lim

n→∞

k=1

k2 a
n3

57. Evaluate:
n

lim 2n

n→∞

k=1

1
1

k(k + 2) 4

n

58. Consider the sequence (an )n≥1 , such that an > 0, (∀)n ∈ N and lim n(an+1 −
n→∞

an ) = 1. Evaluate lim an and lim n an .
n→∞

n→∞

59. Evaluate:

1 + 2 2 + 3 3 + ... + n n

lim
n→∞
n2 n
60. Evaluate:
1

limπ (sin x) 2x−π

x→ 2

61. Evaluate:
lim n2 ln cos

n→∞

1
n

62. Given a, b ∈ R∗+ , evaluate:

n
lim

n→∞

a+
2

63. Let α > β > 0 and the matrices A =

n

1
0

n

b

0
, B=
1

0
1

i)Prove that (∃)(xn )n≥1 , (yn )n≥1 ∈ R such that:
α
β
ii)Evaluate lim

n→∞

β
α

n

= xn A + yn B, (∀)n ≥ 1

xn
.
yn

64. If a ∈ R such that |a| < 1 and p ∈ N∗ is given, evaluate:

1
.
0

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Problems

19

lim np · an

n→∞

65. If p ∈ N∗ , evaluate:
1p + 2p + 3p + . . . + np
n→∞
np+1
lim

66. If n ∈ N∗ , evaluate:
sin(n arccos x)

x→1
1 − x2
lim

x<1

67. If n ∈ N , evaluate:
1 − cos(n arccos x)
1 − x2
x→1
lim

x<1

68. Study the convergence of the sequence:
xn+1 =

xn + a
, n ≥ 1, x1 ≥ 0, a > 0
xn + 1

69. Consider two sequences of real numbers (xn )n≥0 and (yn )n≥0 such that
x0 = y0 = 3, xn = 2xn−1 + yn−1 and yn = 2xn−1 + 3yn−1 , (∀)n ≥ 1. Evaluate
xn
lim
.
n→∞ yn
70. Evaluate:
lim

x→0

tan x − x
x2

71. Evaluate:
lim

x→0

tan x − arctan x
x2

72. Let a > 0 and a sequence of real numbers (xn )n≥0 such that xn ∈ (0, a) and
a2
, (∀)n ∈ N. Prove that (xn )n≥1 is convergent and evaluate
xn+1 (a − xn ) >
4
lim xn .

n→∞

73. Evaluate:
lim cos nπ

n→∞

2n

e

74. Evaluate:

lim

n→∞

n+1
n

tan (n−1)π
2n

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20

A Collection of Limits

75. Evaluate:
n

lim

n

n→∞

k=1

n
k

76. If a > 0, evaluate:
a+

lim

3

a+

n→∞

a + ... +
ln n

n

a−n

77. Evaluate:
π π
+
4
n
78. Let k ∈ N and a0 , a1 , a2 , . . . , ak ∈ R such that a0 + a1 + a2 + . . . + ak = 0.
Evaluate:
lim n ln tan

n→∞

3
a0 3 n + a1 3 n + 1 + . . . + ak n + k

lim

n→∞

79. Evaluate:
lim sin nπ

3

n→∞

n3 + 3n2 + 4n − 5

80. Evaluate:
2 arcsin x − π
sin πx
x→1
lim

x<1

81. Evaluate:
n

lim

n→∞

k=2

1
k ln k

82. Evaluate:

lim  lim

n→∞

1
n3 x2

n

x→0

sin2 (kx)

1+

k=1

83. If p ∈ N , evaluate:
n

lim

n→∞

84. If αn ∈ 0,

π
4

k=0

(k + 1)(k + 2) · . . . · (k + p)
np+1

is a root of the equation tan α + cot α = n, n ≥ 2, evaluate:
lim (sin αn + cos αn )n

n→∞

85. Evaluate:

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Problems

21

n+k
2

n

lim

n→∞

n2

k=1

86. Evaluate:
n

lim

1+

n

n→∞

k=1

k
n

87. Evaluate:
arctan x − arcsin x
x→0
x3
lim

88. If α > 0, evaluate:
(n + 1)α − nα
n→∞
nα−1
lim

89. Evaluate:
n

lim

n→∞

k=1

k2
2k

90. Evaluate:
n

lim

n→∞

k=0

(k + 1)(k + 2)
2k

91. Consider a sequence of real numbers (xn )n≥1 such that x1 ∈ (0, 1) and
xn+1 = x2n − xn + 1, (∀)n ∈ N. Evaluate:
lim (x1 x2 · . . . · xn )

n→∞

92. If n ∈ N∗ , evaluate:
lim

x→0

1 − cos x · cos 2x · . . . · cos nx
x2

93. Consider a sequence of real numbers (xn )n≥1 such that xn is the real root
of the equation x3 + nx − n = 0, n ∈ N∗ . Prove that this sequence is convergent
and find it’s limit.
94. Evaluate:
arctan x − arctan 2
x→2
tan x − tan 2
lim

95. Evaluate:

diendantoanhoc.net [VMF]

22

A Collection of Limits

lim

1+

22

2! +

32

n→∞

3! + . . . +
n

n2

n!

x2
96. Let (xn )n≥1 such that x1 > 0, x1 + x21 < 1 and xn+1 = xn + n2 , (∀)n ≥ 1.
n
1
1
Prove that the sequences (xn )n≥1 and (yn )n≥2 , yn =

are convergent.
xn n − 1
97. Evaluate:
n

lim

n→∞

sin
i=1

2i
n2

98. If a > 0, a = 1, evaluate:
xx − ax
x→a ax − aa
lim

99. Consider a sequence of positive real numbers (an )n≥1 such that an+1 −
1
1
= an +
, (∀)n ≥ 1. Evaluate:
an+1
an
1
1
1
1
lim √
+
+ ... +
n→∞
a2
an
n a1
100. Evaluate:
2arctan x − 2arcsin x
x→0
2tan x − 2sin x
lim

diendantoanhoc.net [VMF]

Chapter 3

Solutions
1. Evaluate:
lim

n→∞

3

n3 + 2n2 + 1 −

3

n3 − 1

Solution:
lim

n→∞

3

n3 + 2n2 + 1 −

3

n3 − 1 = lim

n→∞

n3 + 2n2 + 1 − n3 + 1
3

(n3 + 2n2 + 1)2 +

3

(n3 − 1)(n3 + 2n2 + 1) +
n2 2 +

= lim

n→∞

=

n2

3

1+

2
n

+

1 2
n3

+

3

1−

1
n3

2
3

3
5x + 2 + 2

lim
x→−2
3x + 10 − 2
Solution:

3

5x+10

(5x+2)2 −2 3 5x+2+4
√ 3x+6
3x+10+2

5
lim
3 x→−2
5
=
9

=

3

3x + 10 + 2

(5x + 2)2 − 2 3 5x + 2 + 4

n

3n2 + 9n
, (∀)n ≥ 1.
2
k=1
Prove that this sequence is an arithmetical progression and evaluate:

3. Consider the sequence (an )n≥1 , such that

23

ak =

(n3 − 1)2

2
n

2. Evaluate:

3
5x + 2 + 2
lim √
= lim
x→−2
3x + 10 − 2 x→−2

3

1+

2
n

+

1
n3

+

3

1−

1 2
n3

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